What is M8.4: Newton's third law and equilibrium?

Newton's third-law forces are equal and opposite, act on different bodies and arise from the same interaction; they therefore do not cancel on one body's force diagram.

How many marks are M8.4 questions worth?

M8.4 practice drills on this page are worth 2–7 marks each; marks in a real exam vary by question.

Which past exam papers has M8.4 appeared in?

We have not yet indexed a verified past-paper appearance for M8.4 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

M8.4

Understand and use Newton's third law; equilibrium of forces on a particle and motion in a straight line; apply to smooth pulleys and connected particles; resolve forces in 2 dimensions; equilibrium of a particle under coplanar forces.

Draft — not yet indexed

Newton's third law and equilibrium

Worked answers and methods for M8.4 on Edexcel A-level Maths 9MA0.

Explanation

Newton's third-law forces are equal and opposite, act on different bodies and arise from the same interaction; they therefore do not cancel on one body's force diagram.
For a particle in equilibrium under coplanar forces, resolve in two independent directions and set both component resultants equal to zero.
For connected particles, draw separate force diagrams, use a common acceleration magnitude while the string is taut, and use one tension for a light string over a smooth pulley before solving simultaneous $F=ma$ equations.
A common error is to treat weight and normal reaction as a third-law pair; both act on the same body, whereas a third-law pair acts on different bodies.

Two particles connected by a light string over a smooth pulley share one tension magnitude while the string is taut.

Worked example

A particle is in equilibrium under three coplanar forces. One force is $14\,\text{N}$ east and another is $10\,\text{N}$ at $120^\circ$ anticlockwise from east. Find the third force as a vector in east-north components and find its magnitude.

1.The first two forces have resultant $(14+10\cos120^\circ)\mathbf{i}+(10\sin120^\circ)\mathbf{j}=9\mathbf{i}+5\sqrt3\mathbf{j}$ .
2.Equilibrium requires zero resultant, so the third force is $-9\mathbf{i}-5\sqrt3\mathbf{j}$ .
3.Its magnitude is $\sqrt{9^2+(5\sqrt3)^2}=\sqrt{156}=2\sqrt{39}\,\text{N}$ .

Answer: Third force $=(-9\mathbf{i}-5\sqrt3\mathbf{j})\,\text{N}$ ; Magnitude $=2\sqrt{39}\,\text{N}$

Common mistakes

Don't assign different tension magnitudes to the two ends of one light inextensible string over a smooth pulley.
Don't treat a Newton third-law pair as two forces on the same particle or assume connected-particle tensions without a force diagram.

Exam tip

Draw a separate force diagram for each particle and distinguish interaction pairs from forces that can balance.

Worked practice

Tier 1 · Easy

A book pushes down on a table with force

P

. State the Newton's third-law partner to this force.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	The table pushes upwards on the book with force $P$	2
Notes
The partner must be the same interaction with the bodies reversed. It is the force of the table on the book, equal in magnitude to $P$ and opposite in direction.

(2 marks)

Tier 2 · Standard

Particles of masses

3\,\text{kg}

and

5\,\text{kg}

hang at the ends of a light inextensible string passing over a smooth pulley. Using

g=9.8\,\text{m s}^{-2}

, find the acceleration of the system and the tension in the string.

(5)

(Total for Question 2 is 5 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	Acceleration $2.45\,\text{m s}^{-2}$ , with the $5\,\text{kg}$ particle moving down Tension $36.75\,\text{N}$ , or $36.8\,\text{N}$ to 3 significant figures	5
Notes
Treating both particles as one system, the driving force is $(5-3)g=19.6\,\text{N}$ and the total mass is $8\,\text{kg}$ , so $a=19.6/8=2.45\,\text{m s}^{-2}$ . For the $3\,\text{kg}$ particle moving upward, $T-3g=3a$ , so $T=3(9.8+2.45)=36.75\,\text{N}$ .

(5 marks)

Tier 3 · Hard

4\,\text{kg}

particle on a smooth plane inclined at

30^\circ

is connected by a light inextensible string over a smooth pulley to a freely hanging

3\,\text{kg}

particle. Using

g=9.8\,\text{m s}^{-2}

, find the acceleration of the system and the tension in the string.

(7)

(Total for Question 3 is 7 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	Acceleration $=1.4\,\text{m s}^{-2}$ , with the $3\,\text{kg}$ particle moving down Tension $=25.2\,\text{N}$	7
Notes
The downslope weight component of the $4\,\text{kg}$ particle is $4(9.8)\sin30^\circ=19.6\,\text{N}$ , while the hanging weight is $29.4\,\text{N}$ , so the $3\,\text{kg}$ particle moves down. For the system, $29.4-19.6=7a$ , giving $a=1.4\,\text{m s}^{-2}$ . For the hanging particle, $29.4-T=3(1.4)$ , so $T=25.2\,\text{N}$ .

(7 marks)

Tier 1 · Easy

A tow rope pulls a sledge forwards with force

180\,\text{N}

. State the Newton's third-law partner to this force.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	The sledge pulls the tow rope backwards with force $180\,\text{N}$	2
Notes
The partner force belongs to the same rope-sledge interaction, has equal magnitude and opposite direction, and acts on the other body. It is therefore the $180\,\text{N}$ backward force of the sledge on the rope.

(2 marks)

Tier 2 · Standard

A particle of weight

60\,\text{N}

is held in equilibrium by two light strings. One string extends horizontally to the left from the particle. The other slopes upwards to the right at

45^\circ

to the horizontal. Find the tension in each string.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	Tension in the sloping string $=60\sqrt2\,\text{N}$ Tension in the horizontal string $=60\,\text{N}$	4
Notes
Let the sloping tension be $T$ and the horizontal tension be $H$ . Vertical equilibrium gives $T\sin45^\circ=60$ , so $T=60\sqrt2\,\text{N}$ . Horizontal equilibrium then gives $H=T\cos45^\circ=60\,\text{N}$ .

(4 marks)

Tier 3 · Hard

Particles

A

and

B

, of masses

3\,\text{kg}

and

2\,\text{kg}

respectively, lie on two smooth planes inclined at

30^\circ

and

45^\circ

to the horizontal. They are connected by a light inextensible string passing over a smooth pulley at the common top of the planes. The string lies along a line of greatest slope of each plane. The particles are released from rest. Using

g=9.8\,\text{m s}^{-2}

, find the direction of motion, the acceleration and the tension, giving numerical answers to 3 significant figures.

(6)

(Total for Question 6 is 6 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$A$ moves down the $30^\circ$ plane and $B$ moves up the $45^\circ$ plane Acceleration $=0.168\,\text{m s}^{-2}$ Tension $=14.2\,\text{N}$	6
Notes
The downslope weight components are $3g\sin30^\circ=14.7\,\text{N}$ for $A$ and $2g\sin45^\circ=9.8\sqrt2\,\text{N}$ for $B$ . Since $14.7>9.8\sqrt2$ , $A$ moves down its plane. For the system, $14.7-9.8\sqrt2=5a$ , so $a=0.168\,\text{m s}^{-2}$ to 3 significant figures. For $A$ , $14.7-T=3a$ , giving $T=14.2\,\text{N}$ to 3 significant figures.

(6 marks)

Tier 2 · Standard

Blocks

A

and

B

, of masses

4\,\text{kg}

and

6\,\text{kg}

respectively, are in contact on a smooth horizontal plane. A horizontal force of

30\,\text{N}

is applied to

A

towards

B

. Find the acceleration of the blocks and the force exerted by

A

B

. State the Newton's third-law partner to this contact force.

(5)

(Total for Question 7 is 5 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	Acceleration $=3\,\text{m s}^{-2}$ Force exerted by $A$ on $B=18\,\text{N}$ in the direction of motion The partner is the $18\,\text{N}$ force exerted by $B$ on $A$ in the opposite direction	5
Notes
Treating both blocks as one system, $30=(4+6)a$ , so $a=3\,\text{m s}^{-2}$ . For block $B$ , the only horizontal force is the contact force from $A$ , so it is $6(3)=18\,\text{N}$ in the direction of motion. By Newton's third law, $B$ exerts an $18\,\text{N}$ force on $A$ in the opposite direction.

(5 marks)

Tier 3 · Hard

A particle

A

of mass

5\,\text{kg}

rests on a smooth horizontal table and is connected to a hanging particle

B

of mass

3\,\text{kg}

by a light inextensible string passing over a small smooth pulley at the edge of the table. The string is horizontal between

A

and the pulley. Starting from rest, the particles move with the same acceleration, the string being taut. Take

g=9.8\,\text{m s}^{-2}

. Find their acceleration and the tension in the string. Find their speed after

B

has descended

1.5\,\text{m}

, giving the speed to 3 significant figures. Use unrounded values in your working.

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	Acceleration $=3.675\,\text{m s}^{-2}$ Tension $=18.375\,\text{N}$ Speed $=3.32\,\text{m s}^{-1}$	6
Notes
Let the acceleration be $a$ and the tension be $T$ . For $A$ , $T=5a$ . For $B$ , taking downwards as positive, $3g-T=3a$ . Hence $29.4=8a$ , so $a=3.675\,\text{m s}^{-2}$ and $T=18.375\,\text{N}$ . Using $v^2=u^2+2as$ with $u=0$ and $s=1.5$ , $v=\sqrt{2(3.675)(1.5)}=3.320\ldots\,\text{m s}^{-1}$ , which is $3.32\,\text{m s}^{-1}$ to 3 significant figures.

(6 marks)

Tier 3 · Hard

Particles

A

B

and

C

, of masses

2\,\text{kg}

3\,\text{kg}

and

5\,\text{kg}

respectively, lie in that order on a smooth horizontal plane. Adjacent particles are connected by separate light inextensible strings. A horizontal force of

50\,\text{N}

is applied to

C

away from

B

, and both strings remain taut. Find the acceleration of the particles and the tension in each string. State the Newton's third-law partner to the force exerted by the string joining

B

and

C

on particle

C

. Give all numerical answers exactly.

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	Acceleration $=5\,\text{m s}^{-2}$ Tension between $A$ and $B=10\,\text{N}$ Tension between $B$ and $C=25\,\text{N}$ The partner is the equal and opposite force exerted by $C$ on that string	6
Notes
For the complete system, $50=(2+3+5)a$ , so $a=5\,\text{m s}^{-2}$ . For $A$ , the first tension is the only horizontal force, so $T_1=2(5)=10\,\text{N}$ . The second tension accelerates $A$ and $B$ together, so $T_2=(2+3)(5)=25\,\text{N}$ . The third-law partner to the string's force on $C$ is the force of $C$ on the same string, equal in magnitude and opposite in direction.

(6 marks)

Q10

Tier 3 · Hard

10.

A particle

B

of mass

4\,\text{kg}

lies on a smooth horizontal table. It is connected on its left to a hanging particle

A

of mass

2\,\text{kg}

by one light inextensible string over a smooth pulley, and on its right to a hanging particle

C

of mass

6\,\text{kg}

by a separate light inextensible string over a second smooth pulley. The horizontal parts of both strings are collinear, and both strings remain taut. The system is initially stationary and is then set free. Use

g=9.8\,\text{m s}^{-2}

. Find the direction and exact magnitude of the acceleration, and find the tension in each string. Give all numerical answers exactly.

(7)

(Total for Question 10 is 7 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$C$ moves down, $B$ moves right and $A$ moves up Acceleration $=\dfrac{49}{15}\,\text{m s}^{-2}$ Left-string tension $=\dfrac{392}{15}\,\text{N}$ Right-string tension $=\dfrac{196}{5}\,\text{N}$	7
Notes
The greater hanging weight is that of $C$ , so take $C$ down, $B$ right and $A$ up as positive. For the whole system the driving force is $(6-2)g=39.2\,\text{N}$ and the total mass is $12\,\text{kg}$ , giving $a=39.2/12=49/15\,\text{m s}^{-2}$ . For $A$ , $T_L-2g=2a$ , so $T_L=19.6+98/15=392/15\,\text{N}$ . For $C$ , $6g-T_R=6a$ , so $T_R=58.8-294/15=196/5\,\text{N}$ .

(7 marks)

Verified exam appearances

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Other points in M8 Forces and Newton's laws

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